Problem: You are given N items.
The value of the i-th item (1 \leq i \leq N) is v_i.
Your have to select at least A and at most B of these items.
Under this condition, find the maximum possible arithmetic mean of the values of selected items.
Additionally, find the number of ways to select items so that the mean of the values of selected items is maximized.

Constraints

* 1 \leq N \leq 50
* 1 \leq A,B \leq N
* 1 \leq v_i \leq 10^{15}
* Each v_i is an integer.

Input

The input is given from Standard Input in the following format:


N A B
v_1
v_2
...
v_N


Output

Print two lines.
The first line should contain the maximum possible arithmetic mean of the values of selected items. The output should be considered correct if the absolute or relative error is at most 10^{-6}.
The second line should contain the number of ways to select items so that the mean of the values of selected items is maximized.

Examples

Input

5 2 2
1 2 3 4 5


Output

4.500000
1


Input

4 2 3
10 20 10 10


Output

15.000000
3


Input

5 1 5
1000000000000000 999999999999999 999999999999998 999999999999997 999999999999996


Output

1000000000000000.000000
1

Solution (in JAVA):

Solution:
import java.util.ArrayList;
import java.util.Arrays;
import java.util.Scanner;




class Main{


	static	ArrayList<Integer> one ;
	static	ArrayList<Integer> two ;
	static	ArrayList<Integer> three ;
	static	ArrayList<Integer> four ;

	public static void main(String[] args) {

		Scanner sc = new Scanner(System.in);			//文字の入力
		int n = sc.nextInt();
		int a = sc.nextInt();
		int b = sc.nextInt();
		long[] v =  new long[n];

		for(int i = 0;i < n;i++){
			v[i] = sc.nextLong();
		}
		Arrays.sort(v);
		int j = 0;
		for(int i = 0;i < n;i++){
			if(v[n-1-i] == v[n-1]){
				j++;
			}
		}
		

		long[][] pdp = new long[51][51];
		pdp[0][0] = 1;
		for (int x = 1; x <= 50; x++) {				//組み合わせの数の計算
			for (int k = 0; k <= x; k++) {
				if (k - 1 >= 0) {
					pdp[x][k] = pdp[x-1][k-1] + pdp[x-1][k];
				} else {
					pdp[x][k] = pdp[x-1][k];
				}
			}
		}

		if(j <= a){
			long sum = 0;
			for(int i = 0;i < a;i++){
				sum += v[n-1-i];
			}

			double ans = (double)sum/a;

			int c = 0;

			for(int i = 0;i < n;i++){				//基準数の個数
				if(v[i] == v[n-a]){
					c++;
				}
			}
			int h = 0;
			for(int i = 0;i < n;i++){
				if(v[i] > v[n-a]){
					h++;
				}
			}
			
			
			long ansnum = pdp[c][c-(a-h)];
			


			System.out.println(ans);
			System.out.println(ansnum);
			
		}else{		
			long sum = 0;
			for(int i = 0;i < a;i++){
				sum += v[n-1-i];
			}
			double ans = (double)sum/a;
			
			
			int c = 0;
			for(int i = 0;i < n;i++){				//最大値と同じ値を持つ個数を調べる
				if(v[i] == v[n-1]){
					c++;
				}
			}


			long ansc = 0;
			
			
			for(int i = a;i <= Math.min(c, b);i++){
				ansc += pdp[c][i];
			//System.out.println(pdp[c][i]);

			}
			
			System.out.println(ans);
			System.out.println(ansc);
			
			
			
		}




	}
	
	static long sum(long ue,long sita){
		long counter = sita;
		long u = 1;
		for(long i = 0;i < counter;i++){
			u *= ue;
			ue--;
		}
		long s = 1;
		for(long i = 0;i < counter;i++){
			s *= sita;
			sita--;
		}
		
		return u/s;
		
	}
}

class Pair implements Comparable{
	Character from;
	int end;
	@Override
	public int compareTo(Object other) {
		Pair otherpair = (Pair)other;

		return end - otherpair.end;
	}
}